Monday 14 November 2016

AN INFORMATIVE ARTICLE ON DIRECT FORM OF STRUCTURES IN DSP?

Categories of structures

Structure is the phase of anything in which we put  our focus when we are going to design hardware of  that specific thing.For complete description and demonstration of structures we have two common ways.One is through displaying block diagrams and second one is by writing signal flow graphs of a that specific thing.


direct form I of structures




Filter categorization 

Normally two domains of filters named as analog domain and digital domain cover up all aspects of filters.If we talk about filters generally irrespective of their domains regarding their responses then we reach at the conclusion that finite and infinite impulse responses are two general divisions of filters.If we talk about structural representation of finite and infinite impulse responses filtersthen following forms cover these IIR and FIR filters.

  1. direct form
  2. cascade form
  3. parallel form
  4. transposed for
Types or forms of direct form of structure
  • direct form I
  • direct form II

Features of direct forms

There are interesting and simple differences that are understandable  between direct form I and direct form II of structures.

        H(z) = zeros/poles
where H(z) is called system function

Direct form I
In direct form I we manage the system function in such a way that zeros come first and poles are realized at the second place.
H(z) = (zeros)(poles)
Direct form II
In direct form II system function is expressed in such a way that now poles are realized at first priority and then zeros are realized.
         

         H(z) = (poles)(zeros)

Advantages and disadvantages of direct form of structures

Advantages 
The simple advantage of direct form of structures is its ease of implementation.Furthermore direct form II is superior to direct form I due to the fact that for direct form II  memory locations  is to the least extent whereas direct form I requires more number of memory locations.
disadvantages
Normally direct form of structures is not used so frequently in practical field,the reason is that these structures are affected by the quantization errors in the coefficients.

   
direct form II


Analysis techniques

A variety of mathematical techniques may be employed to analyze the behaviour of a given digital filter. Many of these analysis techniques may also be employed in designs, and often form the basis of a filter specification.

Typically, one characterizes filters by calculating how they will respond to a simple input such as an impulse. One can then extend this information to compute the filter's response to more complex signals.

Impulse response

The impulse response, often denoted {\displaystyle h[k]} h[k] or {\displaystyle h_{k}} h_{k}, is a measurement of how a filter will respond to the Kronecker delta function. For example, given a difference equation, one would set {\displaystyle x_{0}=1} x_{0}=1 and {\displaystyle x_{k}=0} x_{k}=0 for {\displaystyle k\neq 0} k\neq 0 and evaluate. The impulse response is a characterization of the filter's behaviour. Digital filters are typically considered in two categories: infinite impulse response (IIR) and finite impulse response (FIR). In the case of linear time-invariant FIR filters, the impulse response is exactly equal to the sequence of filter coefficients:

{\displaystyle \ y_{n}=\sum _{k=0}^{N}h_{k}x_{n-k}} {\displaystyle \ y_{n}=\sum _{k=0}^{N}h_{k}x_{n-k}}
IIR filters on the other hand are recursive, with the output depending on both current and previous inputs as well as previous outputs. The general form of an IIR filter is thus:

{\displaystyle \ \sum _{m=0}^{M}a_{m}y_{n-m}=\sum _{k=0}^{N}b_{k}x_{n-k}} {\displaystyle \ \sum _{m=0}^{M}a_{m}y_{n-m}=\sum _{k=0}^{N}b_{k}x_{n-k}}
Plotting the impulse response will reveal how a filter will respond to a sudden, momentary disturbance.

Difference equation

In discrete-time systems, the digital filter is often implemented by converting the transfer function to a linear constant-coefficient difference equation (LCCD) via the Z-transform. The discrete frequency-domain transfer function is written as the ratio of two polynomials. For example:

{\displaystyle H(z)={\frac {(z+1)^{2}}{(z-{\frac {1}{2}})(z+{\frac {3}{4}})}}} H(z)={\frac  {(z+1)^{2}}{(z-{\frac  {1}{2}})(z+{\frac  {3}{4}})}}
This is expanded:

{\displaystyle H(z)={\frac {z^{2}+2z+1}{z^{2}+{\frac {1}{4}}z-{\frac {3}{8}}}}} H(z)={\frac  {z^{2}+2z+1}{z^{2}+{\frac  {1}{4}}z-{\frac  {3}{8}}}}
and to make the corresponding filter causal, the numerator and denominator are divided by the highest order of {\displaystyle z} z:

{\displaystyle H(z)={\frac {1+2z^{-1}+z^{-2}}{1+{\frac {1}{4}}z^{-1}-{\frac {3}{8}}z^{-2}}}={\frac {Y(z)}{X(z)}}} H(z)={\frac  {1+2z^{{-1}}+z^{{-2}}}{1+{\frac  {1}{4}}z^{{-1}}-{\frac  {3}{8}}z^{{-2}}}}={\frac  {Y(z)}{X(z)}}
The coefficients of the denominator, {\displaystyle a_{k}} a_{k}, are the 'feed-backward' coefficients and the coefficients of the numerator are the 'feed-forward' coefficients, {\displaystyle b_{k}} b_{{k}}. The resultant linear difference equation is:

{\displaystyle y[n]=-\sum _{k=1}^{M}a_{k}y[n-k]+\sum _{k=0}^{N}b_{k}x[n-k]} y[n]=-\sum _{{k=1}}^{{M}}a_{{k}}y[n-k]+\sum _{{k=0}}^{{N}}b_{{k}}x[n-k]
or, for the example above:

{\displaystyle {\frac {Y(z)}{X(z)}}={\frac {1+2z^{-1}+z^{-2}}{1+{\frac {1}{4}}z^{-1}-{\frac {3}{8}}z^{-2}}}} {\frac  {Y(z)}{X(z)}}={\frac  {1+2z^{{-1}}+z^{{-2}}}{1+{\frac  {1}{4}}z^{{-1}}-{\frac  {3}{8}}z^{{-2}}}}
rearranging terms:

{\displaystyle \Rightarrow (1+{\frac {1}{4}}z^{-1}-{\frac {3}{8}}z^{-2})Y(z)=(1+2z^{-1}+z^{-2})X(z)} \Rightarrow (1+{\frac  {1}{4}}z^{{-1}}-{\frac  {3}{8}}z^{{-2}})Y(z)=(1+2z^{{-1}}+z^{{-2}})X(z)
then by taking the inverse z-transform:

{\displaystyle \Rightarrow y[n]+{\frac {1}{4}}y[n-1]-{\frac {3}{8}}y[n-2]=x[n]+2x[n-1]+x[n-2]} \Rightarrow y[n]+{\frac  {1}{4}}y[n-1]-{\frac  {3}{8}}y[n-2]=x[n]+2x[n-1]+x[n-2]
and finally, by solving for {\displaystyle y[n]} y[n]:

{\displaystyle y[n]=-{\frac {1}{4}}y[n-1]+{\frac {3}{8}}y[n-2]+x[n]+2x[n-1]+x[n-2]} y[n]=-{\frac  {1}{4}}y[n-1]+{\frac  {3}{8}}y[n-2]+x[n]+2x[n-1]+x[n-2]
This equation shows how to compute the next output sample, {\displaystyle y[n]} y[n], in terms of the past outputs, {\displaystyle y[n-p]} y[n-p], the present input, {\displaystyle x[n]} x[n], and the past inputs, {\displaystyle x[n-p]} x[n-p]. Applying the filter to an input in this form is equivalent to a Direct Form I or II realization, depending on the exact order of evaluation.Courtesy of wikipedia,,,




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